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g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
Tìm x :
a) \(x-5=49:7\)
\(\Leftrightarrow x-5=7\)
\(\Leftrightarrow x=7+5\)
\(\Leftrightarrow x=12\)
Vậy : \(x=12\)
b) \(2x+6=24\)
\(\Leftrightarrow2x=24-6=18\)
\(\Leftrightarrow x=18:2\)
\(\Leftrightarrow x=9\)
Vậy : \(x=9\)
c) \(\frac{1}{3}:x+\frac{1}{2}=5\)
\(\Leftrightarrow\frac{1}{3}:x=5-\frac{1}{2}=\frac{9}{5}\)
\(\Leftrightarrow x=\frac{1}{3}:\frac{9}{5}\)
\(\Leftrightarrow x=\frac{5}{27}\)
Vậy : \(x=\frac{5}{27}\)
d) \(\frac{1}{6}.x-\frac{1}{3}=2\)
\(\Leftrightarrow\frac{1}{6}.x=2-\frac{1}{3}=\frac{5}{3}\)
\(\Leftrightarrow x=\frac{5}{3}:\frac{1}{6}\)
\(\Leftrightarrow x=10\)
Vậy : \(x=10\)
e) \(\frac{x}{27}=\frac{3}{x}\)
\(\Leftrightarrow x.x=27.3\)
\(\Leftrightarrow x^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=9\end{matrix}\right.\) mà \(x\in N\)
\(\Rightarrow x=9\)
Vậy : \(x=9\)
g) \(1200:24-\left(17-x\right)=36\)
\(\Leftrightarrow50-17+x=36\)
\(\Leftrightarrow33+x=36\)
\(\Leftrightarrow x=36-33\)
\(\Leftrightarrow x=3\)
Vậy : \(x=3\)
h) \(674-\left(12+x\right)=427\)
\(\Leftrightarrow12+x=674-427=247\)
\(\Leftrightarrow x=247-12\)
\(\Leftrightarrow x=230\)
Vậy : \(x=230\)
k) \(36.\left(x-9\right)=900\)
\(\Leftrightarrow x-9=900:36\)
\(\Leftrightarrow x-9=25\)
\(\Leftrightarrow x=25+9\)
\(\Leftrightarrow x=34\)
m) \(1,2:x+3,8:x=2,5\)
\(\Leftrightarrow\left(1,2-3,8\right):x=2,5\)
\(\Leftrightarrow-2,6:x=2,5\)
\(\Leftrightarrow x=\frac{-2,6}{2,5}=-\frac{26}{25}\)
Vậy : \(x=-\frac{26}{25}\)
n) \(\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right).x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right).x=\frac{1}{3}\)
\(\Leftrightarrow\left[\frac{1}{2}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\right].x=\frac{1}{3}\)
\(\Leftrightarrow\left[\frac{1}{2}.\left(1-\frac{1}{10}\right)\right].x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}.\frac{9}{10}.x=\frac{1}{3}\)
\(\Leftrightarrow\frac{9}{20}.x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}:\frac{9}{20}\)
\(\Leftrightarrow x=\frac{20}{27}\)
Vậy : \(x=\frac{20}{27}\)
a)\(\frac{5}{6}-x=-\frac{7}{12}+\frac{2}{3}\)
\(\frac{5}{6}-x=\frac{1}{12}\)
\(x=\frac{5}{6}-\frac{1}{12}\)
\(\Rightarrow x=\frac{3}{4}\)
b)\(\left(2,4x-36\right):1\frac{5}{7}=-14\)
\(\left(2,4x-36\right)=-24\)
\(2,4x=12\)
\(\Rightarrow x=5\)
c)\(\left(3\frac{1}{2}+2x\right).3\frac{2}{3}=5\frac{1}{3}\)
\(3\frac{1}{2}+2x=\frac{16}{11}\)
\(2x=-\frac{45}{22}\)
\(x=-\frac{45}{44}\)
d)\(\frac{5}{6}-\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{3}{8}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{11}{24}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{11}{24}\\\frac{1}{2}x-\frac{1}{3}=-\frac{11}{24}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{19}{12}\\x=-\frac{1}{4}\end{cases}}\)
e)\(\left|\frac{1}{4}-2x\right|-\frac{3}{4}=0\)
\(\left|\frac{1}{4}-2x\right|=\frac{3}{4}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{4}-2x=\frac{3}{4}\\\frac{1}{4}-2x=-\frac{3}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{4}\\x=\frac{1}{2}\end{cases}}\)
Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
a)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\Leftrightarrow x\left(x-1\right)^{x+2}\left(x-2\right)=0\)
Do đó \(x\in\left\{0;1;2\right\}\)
b)
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot...\cdot\frac{31}{64}=2^x\Leftrightarrow\frac{1\cdot2\cdot3\cdot...\cdot31}{4\cdot6\cdot8\cdot...\cdot64}=2^x\Leftrightarrow\frac{31!}{\left(2\cdot2\right)\cdot\left(2\cdot3\right)\cdot\left(2\cdot4\right)\cdot...\cdot\left(2\cdot31\right)\cdot64}=2^x\)
\(\frac{31!}{2^{30}\cdot31!\cdot2^6}=2^x\Leftrightarrow\frac{1}{2^{36}}=2^x\Leftrightarrow2^{-36}=2^x\Rightarrow x=-36\)
Ta có: \(\frac{12}{36}=\frac{1}{3}\)mà \(\frac{12}{36}=\frac{2}{3\times x}\)
\(\Rightarrow\frac{1}{3}=\frac{2}{3\times x}\)
Vậy x = (2 : 1) x 3 = 6
Vậy x = 6
Mấy câu kia TT